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%I #25 Mar 23 2023 03:31:18
%S 77,311,701,1247,1949,2807,3821,4991,6317,7799,9437,11231,13181,15287,
%T 17549,19967,22541,25271,28157,31199,34397,37751,41261,44927,48749,
%U 52727,56861,61151,65597,70199,74957,79871,84941,90167,95549,101087,106781,112631,118637
%N a(n) = 78*n^2 - 1.
%C The identity (78*n^2 - 1)^2 - (1521*n^2 - 39)*(2*n)^2 = 1 can be written as a(n)^2 - A158770(n)*A005843(n)^2 = 1.
%H Vincenzo Librandi, <a href="/A158771/b158771.txt">Table of n, a(n) for n = 1..10000</a>
%H Vincenzo Librandi, <a href="https://web.archive.org/web/20090309225914/http://mathforum.org/kb/message.jspa?messageID=5785989&tstart=0">X^2-AY^2=1</a>, Math Forum, 2007. [Wayback Machine link]
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F G.f.: x*(-77 - 80*x + x^2)/(x-1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F From _Amiram Eldar_, Mar 23 2023: (Start)
%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/sqrt(78))*Pi/sqrt(78))/2.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/sqrt(78))*Pi/sqrt(78) - 1)/2. (End)
%t LinearRecurrence[{3, -3, 1}, {77, 311, 701}, 50] (* _Vincenzo Librandi_, Feb 21 2012 *)
%t 78*Range[40]^2-1 (* _Harvey P. Dale_, Nov 28 2018 *)
%o (Magma) I:=[77, 311, 701]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // _Vincenzo Librandi_, Feb 21 2012
%o (PARI) for(n=1, 40, print1(78*n^2 - 1", ")); \\ _Vincenzo Librandi_, Feb 21 2012
%Y Cf. A005843, A158770.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 26 2009
%E Comment rewritten and formula replaced by _R. J. Mathar_, Oct 22 2009
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